In [1]:

```
from __future__ import division
import math
# given data:
p=8; # number of poles
a1=p; # in lap winding
a2=2; # in wave winding
fi=15*10**-3;# in wb
N=500;# rev/min
Z=800;# number of conductors on armature
#calculations:
emf1=(fi*Z*N*p)/(60*a1)# when the armature is lap wound
emf2=(fi*Z*N*p)/(60*a2)# when the armature is wave wound
#Results
print "when the armature is lap wound, emf(V) = ",emf1
print "when the armature is wave wound, emf(V) = ",emf2
```

In [2]:

```
from __future__ import division
import math
#given data:
Vt=200# terminal voltage in volts
Rsh=100;#shunt fieldresistance in ohm
Ra=0.1;# armature resistance in ohm
l=60;# number of lamps
w=40 # in watt
N=4; # number of poles
#calculations:
total_l=l*w# in watt
Il=total_l/Vt# load current
Ish=Vt/Rsh# shunt field current
Ia=Il+Ish;
I=Ia/N;
Va=Ia*Ra#armature voltage drop
Vb=1+1;# brush contact drop for 2 pair of poles
E=Vt+Va+Vb;
#Results
print "(a)armature current,Ia(A) = ",Ia
print "(b)current per path in a armature,I(A) =",I
print "(c)emf,E(Volts) = ",E
```

In [3]:

```
from __future__ import division
import math
# given data:
W=10 # output of the generator in k-w
V=250;# voltage in volts
R=0.07;# in ohm
Rsh=63.2;# shunt resistance in ohm
Ra=0.05;# armature resistance in ohm
Vb=2;# brush contact drop
#calculations:
Il=(W*1000)/V# load current in A
Vf=Il*R# voltage drop in feeder
Vt=V+Vf;
Ish=Vt/Rsh;
Ia=Il+Ish;
Vd=Ia*Ra# voltage drop in the armature
E=Vt+Vd+Vb;
#Results
print "(a)terminal voltage,Vt(V) = ",Vt
print "(b)emf,E(V) = ", E
```

In [4]:

```
from __future__ import division
import math
# given data:
W=20000# in watt
V=200;# in volts
R=0.08;# in ohm
Rs=0.02;# series field resistance in ohm
Rsh=42;# shunt ield resistance in ohm
Ra=0.04;# armature resistance in ohm
iron_losses=309.5;# iron and friction losses
#calculations:
I=W/V;# in A
Vf=I*R;
Vs=I*Rs;
V1=Vf+Vs;# voltage drop of feeder and series field
Vg=V+V1;
Ish=Vg/Rsh# shunt field current
Ia=I+Ish;
Vd=Ia*Ra;
emf=Vg+Vd;
Ed=emf*Ia# in watt
copper_losses=Ed-W;
mech_in=W+copper_losses+iron_losses;
Bhp=mech_in/735.5;
efficiency=(W/mech_in)*100;
#Results
print "(a)terminal voltage,Vg(V) = ",Vg
print "(b)emf(V) =",emf
print "(c)copper losses(Watt) = ",copper_losses
print "(d)bhp metric of the primemover,Bhp = ",Bhp
print "(e)efficiency(%) = ",round(efficiency,1)
```

In [5]:

```
from __future__ import division
import math
# given data:
n=3 # number of motors
n1=4 # number of parallel path in winding
i=30;#current in A
Bhp=65# in hp
Rsh=44;# shunt field resistance
Ra=0.08;# armature resistance in ohm
V=440;# voltage in V
Vb=2 # we know , brush contact drops
#calculations:
I=i*n# current taken by three motors
Ish=V/Rsh# shunt field current
Ia=I+Ish;
I1=Ia/n1# current in each path
Va=Ia*Ra;# armature drop
E=V+Va+Vb;
E_power=E*Ia;
W=V*I# in watt
M_power=Bhp*746# assume Bhp=746 W
Copper_losses=E_power-W;
S_loses=M_power-E_power;
eta_e=(W/E_power)*100;
eta_c=(W/M_power)*100;
eta_m=(E_power/M_power)*100;
#Results
print "(a)total armature current,Ia(A) =",Ia
print "(b)current in each path,I1(A) = ",I1
print "(c)emf,E(V) = ",E # answer is wrong in a book
print "(d)electrical power developed in watt = ",E_power # answer is wrong in a book
print "(e)copper losses (W) = ",Copper_losses
print "(f)stray losses(W) = ",S_loses
print "(g1)electrical efficiency,eta_e(%) = ",eta_e
print "(g2)commercial efficiency,eta_c(%) = ",round(eta_c,2)
print "(g3)mechanical efficiency,eta_m(%) = ",round(eta_m,1)
```